Intriguing Limit

Date: 04/30/97 at 19:33:57
From: oliver
Subject: (1-1/x)^x
I am curious to see what chance you have of winning a contest if you
have a 1 in 10 chance of winning each time you play and you play 10
times. I computed this as follows: chance = 1-.9^10 = .65... What
would it be for a 1 in 20 chance if you play 20 times? How about a 1
in 100 chance when you play 100 times? And so on.
The function then looks like y = 1-(1-1/x)^x, where x is the chance
you have of winning and the number of times you play. So if x = 100
you have a 1 in 100 chance of winning each time you play and you play
100 times. y is then the chance that you will win at least once.
What is the limit of this as x->infinity? When I graph it, it looks
asymptotic at about .632..., but the function looks to me like it has
a limit of 0. (1/x goes to 0 and 1 to any power is one.) I asked a
professor about this, and he told me that (1-1/x)^x has a special
limit, namely 1/e. This then gives the expected result.
My question is: Why is lim (1-1/x)^x = 1/e?
Thank you,
Oliver Dale

Date: 05/01/97 at 09:01:05
From: Doctor Jerry
Subject: Re: (1-1/x)^x
Hi Oliver,
There are several different reasons "why" this limit has this
particular value. I'll give one answer below, based on a calculation
using l'Hopital's Rule, from calculus. If you're asking why is this
limit the number 2.71828..., the base of the natural logarithms, the
answer is that this limit comes up in working with exponentials and
logarithms and was long ago given the name e, after Euler. Usually,
it comes up in the form (1+1/n)^n, which approaches e as n->oo.
Let y = (1-1/x)^x. Take natural logs of both sides.
ln(y) = x*ln(1-1/x). As x->oo, this has the form of oo*0 and so must
be rearranged. We have:
ln(y) = ln(1-1/x)/(1/x)
Now the form is 0/0 and so l'Hopital's Rule is applicable. After
differentiating numerator and denominator separately (according to
l'Hopital's Rule) and simplifying (I'll use lim to mean limit as x->
oo), we have:
lim ln(y) = lim (-1)/(1-1/x) = -1.
So ln(y) -> -1. This means that y->e^(-1).
-Doctor Jerry, The Math Forum
Check out our web site! http://mathforum.org/dr.math/